The millwright & engineer's pocket companion

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Simpkin, Marshall and Company, 1846 - 220 ページ
 

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91 ページ - ... one of the most important, and at the same time, one of the least expensive and troublesome, which we possess.
74 ページ - The areas of circles are to each other as the squares of their diameters.
36 ページ - Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
83 ページ - To find the solidity of a spheroid. RULE. Multiply the square of the revolving axis by the fixed axis, and by *5236, and the product will be the solidity.
74 ページ - Area of a circle is equal to the area of a triangle whose base equals the circumference and perpendicular equals the radius.
104 ページ - These simple machines are the lever, the wheel and axle, the pulley, the inclined plane, the wedge, and the screw.
69 ページ - NOTE. — 1. As 7 is to 22, so is the diameter to the circumference; or, as 22 is to 7, so is the circumference to the diameter.
104 ページ - ... that there may be a balance between the power and the weight, the intensity of the power must exceed the intensity of the weight just as much as the distance of the weight from the prop exceeds the distance of the power.
36 ページ - ... and to the remainder bring down the next period for a dividend. 3. Place the double of the root already found, on the left hand of the dividend for a divisor. 4. Seek how often the divisor is contained...
84 ページ - ... by •5236 ; then say, as the square of the fixed axis is to the square of the revolving axis, so is the former product to the solidity.

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